Difference between revisions of "2012 AMC 12B Problems/Problem 25"

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Therefore <math>\boxed{\textbf{(B)}  \ \frac{625}{144}}</math> is the answer.
Therefore <math>\boxed{\textbf{(B)}  \ \frac{625}{144}}</math> is the answer.
== See Also ==
{{AMC12 box|year=2012|ab=B|num-b=24|after=}}

Revision as of 22:50, 12 January 2013

Problem 25

Let $S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}$. Let $T$ be the set of all right triangles whose vertices are in $S$. For every right triangle $t=\triangle{ABC}$ with vertices $A$, $B$, and $C$ in counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. What is \[\prod_{t\in T} f(t)?\]

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24}$


Solution 1

Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that $\tan x  \tan(90^{\circ}-x) = \tan x  \cot x = 1.$

Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.

So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangles $\text{tan B} = \frac{y}{x}$. Multiplying them all together gives: $\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot  \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot  \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot  \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot  \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. }$

Solution 2

Consider reflections. For any right triangle $ABC$ with the right labeling described in the problem, any reflection $A'B'C'$ labeled that way will give us $\tan CBA \cdot \tan C'B'A' = 1$. First we consider the reflection about the line $y=2.5$. Only those triangles $\subseteq S$ that have one vertex at $(0,5)$ do not reflect to a traingle $\subseteq S$. Within those triangles, consider a reflection about the line $y=5-x$. Then only those triangles $\subseteq S$ that have one vertex on the line $y=0$ do not reflect to a triangle $\subseteq S$. So we only need to look at right triangles that have vertices $(0,5), (*,0), (*,*)$. There are three cases:

Case 1: $A=(0,5)$. Then $B=(*,0)$ is impossible.

Case 2: $B=(0,5)$. Then we look for $A=(x,y)$ such that $\angle BAC=90^{\circ}$ and that $C=(*,0)$. They are: $(A=(x,5), C=(x,0))$, $(A=(2,4), C=(1,0))$ and $(A=(4,1), C=(3,0))$. The product of their values of $\tan \angle CBA$ is $\frac{5}{1}\cdot  \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}$.

Case 3: $C=(0,5)$. Then $A=(*,0)$ is impossible.

Therefore $\boxed{\textbf{(B)}  \ \frac{625}{144}}$ is the answer.

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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